100 cc of N/10 NaOH solution is mixed with 100 cc of N/5 HCl solution and the whole volume is made to 1 litre.The pH of the resulting solution will be??

NaOH  volume = 100ml or 0.1L  normality =N/10  molarity=normality= 0.1M
moles of NaOH =  0.1 $\times$ 0.1 = 0.01mol  [using formula  mole = C (concentration in M)  $\times$ V in L]

HCl  volume = 100ml or 0.1L  normality = molairty = 0.2M
Moles of HCl = 0.2  $\times$ 0.1 = 0.02mol

0.01 mole of NaOH neutralizes 0.01 mole of HCl
As HCl is in excess, moles of HCl left unreacted = 0.02-0.01 = 0.01mol
These free ions determine the pH of the solution
Moles of HCl left after neutralization = 0.01mol
Total volume of HCl solution = 1L
Concentration of HCl solution = moles of HCl/ volume of solution in L  = 0.01M
pH = -log[H⁺]
pH = -(log 0.01) =  -(-2) = 2